Using the fact that $\mathbb{R}^n $ is separable, show that every open cover of $\mathbb{R}^n$ has a countable subcover.

I was given a hint to prove the following claim and if needed, incorporate it into the original proof: ''A set is open if and only if it is equal to an arbitrary collection of open balls", which I have been able to prove both ways. However, I don't see how I can use that to go about the original proof.

Otherwise, I thought of the following statements
$(i)$ $\mathbb{R}^n$ is separable
$(ii)$ $\mathbb{R}^n$ is compact
$(iii)$ Any open cover of $\mathbb{R}^n$ has a countable subcover

It would have been great if I could have proved $(i)\Rightarrow(ii)\Rightarrow(iii)$ but the problems that arise are
1. Firstly, $\mathbb{R}^n$ is not compact. Also, ''Every compact metric space is separable", which means $(ii)\Rightarrow(i)$.
2. ''For a compact space, every open cover has a finite subcover", which is not equivalent to countability.

How should I prove this? How are separability and claimed fact requirements to the proof?

Edit. The attempted second method is obviously wrong because $\mathbb{R}^n$ is not compact.


2 Answers 2


Cover $\Bbb R^n$ with any countable collection of compact sets (e.g. all closed balls of radius $\sqrt n$ centred on points with integral coordinates). Now just cover each of these sets separately with a finite subcover. A countable union of finite covers is countable.

I don't see where I've used the separability of $\Bbb R^n$, but I'm sure it's hidden somewhere in there...

  • 1
    $\begingroup$ Compact subsets of \Bbb R^n are separable so a countable union of compact subsets of \Bbb R^n is separable. $\endgroup$ Dec 3, 2020 at 19:59
  • $\begingroup$ I think it's fair to say you haven't used the separability of $\mathbb{R}^n$. Rather what you've done is show that every $\sigma$-compact space is has the "Lindelöf property" (every open cover admits a countable subcover). Even a compact space doesn't need to be separable. $\endgroup$
    – Mike F
    Dec 3, 2020 at 20:32

$\mathbb R^n = \bigcup_{x \in \mathbb Z^n} ([0,1]^n + x)$, write $A_x = [0,1]^n + x$. The $A_x$ cover $\mathbb R^n$. Let $\mathcal U$ be a cover of $\mathbb R^n$, then $\mathcal U$ covers $A_x$ and has a finite subcover $\mathcal U_x$. The set $\bigcup_{x \in \mathbb Z^n} \mathcal U_x$ is a countable cover.

Here are generalizations, as mentioned in a comment:

  1. Every hemicompact space is countably compact.
  2. Every separable metric space is countably compact.
  • 1
    $\begingroup$ Separable metrizable spaces are 2nd-countable, and 2nd-countable spaces are Lindelof. $\endgroup$ Dec 3, 2020 at 20:01

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